Representing and Analyzing SHM Practice

Sample questions

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1. Sample 1difficulty 2/5

   The graph shows $v(t)$ for a mass-on-spring oscillator. At which instant is the mass at one of the <strong>amplitude positions</strong> (largest displacement)?

   • A

     $t = 3T/4$

   • B

     $t = 0$

   • C

     $t = T/2$

   • D

     $t = T/4$

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   Why

   At the amplitude (turning points), velocity equals zero. On the graph the first zero after $t = 0$ is at $t = T/4$ (where $v$ is at a minimum of the cosine — actually $v$ starts from $0$ going negative; zero at $T/4$ corresponds to max negative displacement). Any zero-$v$ instant works; $T/4$ is one such time.

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2. Sample 2difficulty 2/5

   A mass $m = 0.25~\text{kg}$ on a spring ($k = 100~\text{N/m}$) oscillates with amplitude $A = 0.10~\text{m}$. What is its maximum speed?

   • A

     $0.40~\text{m/s}$

   • B

     $1.0~\text{m/s}$

   • C

     $4.0~\text{m/s}$

   • D

     $2.0~\text{m/s}$

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   Why

   Energy: $\tfrac{1}{2}k A^2 = \tfrac{1}{2}m v_{\max}^2$ $\Rightarrow v_{\max} = A\sqrt{k/m} = 0.10\sqrt{100/0.25} = 0.10(20) = 2.0~\text{m/s}$.

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3. Sample 3difficulty 2/5

   A mass oscillates as $x(t) = A\cos(\omega t)$. At $t = T/4$, where is it?

   • A

     $x = 0$

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   • B

     $x = +A/2$

   • C

     $x = +A$

   • D

     $x = -A$

   Why

   $\omega T = 2\pi$, so $\omega t = \pi/2$ at $t = T/4$. $\cos(\pi/2) = 0$.

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4. Sample 4difficulty 2/5

   A mass starts at $x = +A$ with zero velocity at $t = 0$. The motion is described by

   • A

     $A\cos(\omega t)$

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   • B

     $-A\sin(\omega t)$

   • C

     $A\sin(\omega t)$

   • D

     $A\sin(\omega t + \pi/4)$

   Why

   At $t = 0$: $x = A$, $v = 0$. The cosine form satisfies both conditions: $A\cos 0 = A$ and $-A\omega\sin 0 = 0$.

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5. Sample 5difficulty 2/5

   For the oscillator above ($m = 0.25~\text{kg}$, $k = 100~\text{N/m}$, $A = 0.10~\text{m}$), what is the maximum acceleration?

   • A

     $40~\text{m/s}^2$

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   • B

     $10~\text{m/s}^2$

   • C

     $25~\text{m/s}^2$

   • D

     $4.0~\text{m/s}^2$

   Why

   $a_{\max}$ occurs at extremes where force is greatest: $a_{\max} = k A / m = (100)(0.10)/0.25 = 40~\text{m/s}^2$.

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